Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560
[tex]P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})[/tex]
[tex]P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})[/tex]
[tex]P(X > 1560) = P(Z > \dfrac{110}{220})[/tex]
P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean [tex]\mu_x[/tex] = np = 52 × 0.3085 = 16.042
The standard deviation = [tex]\sqrt {n \time p (1-p)}[/tex]
The standard deviation = [tex]\sqrt {52 \times 0.3085 (1-0.3085)}[/tex]
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)
[tex]Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})[/tex]
[tex]Pr ( Y > 20) = P(Z >1 .338)[/tex]
From z tables
P(Y > 20) [tex]\simeq[/tex] 0.0903
